A series circuit has a 50-Hz ac source, a 50-ohm resistor, a 0.50-H inductor, an

Question

A series circuit has a 50-Hz ac source, a 50-ohm resistor, a 0.50-H inductor, and an 60-F capacitor, as shown. The rms current in the circuit is 3.1 A.

A)The voltage amplitude of the source is?
B)The ms voltage across points a and b in the circuit is?
C)The frequency of the source is changed so that the capacitive reactance is equal to twice the inductive reactance. The original circuit elements are retained. The new frequency of the source, in Hz, is?
D)The capacitance is changed so that the circuit is in
resonance. The original resistor and inductor are retained. The voltage and frequency of the source are kept at the original values. The new capacitance, in F is?

Explanation / Answer

Given that, frequency of the A.C circuit f = 50 Hz inductance of the inductor L = 0.5 H the resistance R = 50 the capacity of the capacitor C = 60 F                                                = (60 F)(10-6 F/1 F)                                                = 60*10-6 F the rms current in the circuit Irms = 3.1 A the inductive reactance XL = 2fL                                          = (2)(50 Hz)(0.5 H)                                          = 157 the capacitive reactance XC = 1/2fC                                           = 1/[ (2)(50 Hz)(60*10-6 F)                                           = 53.078 _________________________________________________ a) The rms voltage of the source is           Vrms = IrmsZ here, the circuit impedance Z = R2+(XL-XC)2                                             = (50 )2+(157 -53.078 )2                                             = 115.32 therefore, the rms voltage is           Vrms = (3.1 A)(115.32 )                    = 357.5 V The voltage amplitude of the source is                V = Vrms2                    = (357.5 V)2                    = 505.58 V _____________________________________________________ The voltage amplitude of the source is                V = Vrms2                    = (357.5 V)2                    = 505.58 V _____________________________________________________ b) voltage across resistor is                            VR = IrmsR                                  = (3.1 A)(50 )                                  = 155 V voltage across inductor is                            VL = IrmsXL                                  = (3.1 A)(157 )                                  = 486.7 V the voltage across resistor and capacitor = 155 V + 486.7 V                                                              = 641.7 V therefore, the rms voltage across points a and b in the circuit is            Vrms = V/2                    = (641.7 V)/2                    = 453.75 V ____________________________________________________ ____________________________________________________ c) in this case,                  XC = 2XL            (1/2fC) = 2(2fL) therefore, the new frequency is                     f = [1/82CL]1/2                          = [1/82(60*10-6 F)(0.5 H)]1/2                          = [1/82(60*10-6 F)(0.5 H)]1/2                          = 20.557 Hz ____________________________________________________ ____________________________________________________ d) at resonance condition, XC = XL                              (1/2fC) = (2fL) the new capacitance is                            C = 1/42f2L                               = 1/ 42(50 Hz)2(0.5 H)                               = 20.28*10-6 F                               = 20.28 F   

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